# DC branch current analysis

Discussion in 'General Electronics Chat' started by piracyer, Jun 24, 2012.

Jun 9, 2012
32
1
2. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
I have worked through the calculation for I1 below.

Code ( (Unknown Language)):
1.
2. $$I_{1}\;+\;I_{2}\;-\;0.005\; =\; 0$$
3.
4. $$\frac{6\;-\;V_{x}}{1000}\;+\;\frac{7.2\;-\;V_{x}}{1000}\;-\;0.005\;=\;0$$
5.
6. $${6\;-\;V_{x}}\;+\;{7.2\;-\;V_{x}}\;-\;5\;=\;0$$
7.
8. $${-2V_{X}\;+\;(7.2+6)\;-\;5\;=\;0$$
9.
10. $${-2V_{X}\;+\;8.2\;=\;0$$
11.
12. $${2V_{X}\;=\;8.2$$
13.
14. [TEX]V_{x}\;=\frac{8.2}{2}[/TEX]
15.
16. [TEX]V_{x}\;=\;4.1[/TEX]
17.
18. Plugging the value for Vx back into the expression for the I1:
19.
20. $$I_{1}\;=\;\frac{6\;-\;4.1}{1000}$$
21.
22. $$I_{1}\;=\;\frac{1.9}{1000}$$
23.
24. $$I_{1}\;=\;{1.9}\; milliamps$$
25.
hgmjr

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3. ### piracyer Thread Starter Member

Jun 9, 2012
32
1
great, thank you hgmjr!

I tried solve the problem with KVL, what's wrong with my solution below?

let "I" be the current from the 6v battery and "Vx" be the voltage drop at the transistor.

1000I + Vx + 6 = 0 (1)

1000(0.005 - I) + Vx + 7.2 = 0
Vx - 1000I + 12.2 = 0 (2)

(1) - (2) -> 2000I - 6.2 = 0
I = 0.0036

1000(0.0036) + Vx + 6 = 0
Vx = -9.6

I know the results from my solution looks very wrong since the voltage drop at the transistor is higher than both power source. However, can someone help my identify where exactly I have made wrong in the solution?

Thanks.

4. ### WBahn Moderator

Mar 31, 2012
23,157
6,972
I think one of the points of this particular problem, as indicated by the hint, is that the solution can often be streamlined by leveraging knowledge of how the unknowns interact.

In this case, we know that if the current we want, namely that from the 6V battery, is I, that the current from the 7.2V battery is (5mA-I) by means of applying KCL at the T junction by inspection.

We can then proceed to write expressions for the voltage at the T coming from the two directions and set them equal as follows:

$
6V \ - \ I1k\Omega \ = \ 7.2V \ - \ (5mA-I)1k \Omega
\
\frac{6V}{1k\Omega} \ - \ I \ = \ \frac{7.2V}{1k\Omega} \ - \ (5mA-I)
\
6mA \ - \ I \ = \ 7.2mA \ - \ 5mA \ + \ I \ = \ 2.2mA \ + \ I
\
2I \ = \ 6mA - 2.2mA = \ 3.8mA
\
I \ = \ 1.9mA
$

Notice that, approached this way, you can actually work the problem in your head fairly handily.

Having said that, I admit that I tend to approach things in a more uniform and consistent method even if it is pretty evident that a particular problem has a more straightforward approach. The reason is simple -- we get good at the things we do regularly and repeatedly. But learning to spot them is still good practice and it lets you sanity check your answers better if you can do some quick mental estimates.

For instance, in this case you come up with 1.9mA. So let's round that to 2mA which means that the voltage at the T coming from the 6V battery is 4V. The current in the 7.2V battery must be 3mA, so the voltage at the T coming from that direction is 4.2V. Those voltages are close enough to be consistent with the estimate, so the answer is either correct or very nearly so. If, on the other hand, we had come up with 3.8mA (forgot to divide by 2), the we might estimate the answer as 4mA and then gotten 2V from the left and 5.2V from the right. Those are far enough apart that we know we did something wrong.

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5. ### WBahn Moderator

Mar 31, 2012
23,157
6,972
You have to apply KVL consistently as you make a complete circuit through the network. If you cross a voltage source from negative to positive, you add that voltage otherwise you subtract it. If you cross a passive component (resistor, in this case) in the direction of the current you are assuming, you subtract the voltage otherwise you add it. If you move from a voltage as Node 1 to a voltage at Node 2, you subtract (V1 - V2).

Since you are saying that I is the current 'from' the 6V battery, that means it is moving from left to right through the resistor. So, making a clockwise loop starting from the bottom of the battery, we first gain 6V, then drop (1000Ω)I. then drop from Vx back to ground. So our loop equation is:

6V - (1000Ω)I - Vx = 0

Similarly, if the current coming out of the 7.2V battery is (0.005A - I), then a loop going counterclockwise yields:

7.2V - (1000Ω)(0.005A - I) - Vx = 0

Now subtract the first from the second to get:

1.2V - (1000Ω)[(0.005A - I) - I] = 0

1.2V = (1000Ω)(0.005A - 2I)

1.2V/(1000Ω) = 1.2mA = 0.005A - 2I

2I = 0.005A - 1.2mA = 5mA - 1.2mA = 3.8mA

I = 1.9mA

You can chose which direction you go around a given loop and you can choose whether you are going to add voltage drops or voltage gains as you go around it. But whatever you choose, you must then apply it consistently. You can make different choices on the next loop, but on each loop you must apply the choices made for that loop consistently.

Also, do yourself a really, really big favor. Start using units consistently and religiously throughout your work from beginning to end on every line. Tracking units is perhaps the single most valuable error detecting tool you have, because most (not all, but most) errors you make will screw up the units. But you can only detect those errors if the units are there to get screwed up.

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6. ### piracyer Thread Starter Member

Jun 9, 2012
32
1
Thank you so much, that's very helpful!

7. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
I'm a big Millman's Theorem fan so I go at the problem from that direction.

Code ( (Unknown Language)):
1.
2.
3. $$V_{x}\;=\;\frac{\frac{6}{1000}\;+\;\frac{7.2}{1000}\;-\;0.005}{\frac{1}{1000}\;+\;\frac{1}{1000}}$$
4.
5.
6. $$\;$$
7.
8.
With Vx calculated, I have all of the information I need to determine the current in the lefthand 1K resistor.

hgmjr

8. ### piracyer Thread Starter Member

Jun 9, 2012
32
1
That's a nice theorem, looks very handy.

However, I still have a question regarding the current analysis. When I am doing mesh current analysis, I found it's much easier to use conventional current flow rather than the flow of electrons, even though I don't have any prior electrical knowledge. I figured that is probably because the signs of batteries make the flow of electrons very unintuitive. However, is there any trick in using the flow of electrons for current analysis? (I kind of wanna learn this way because it's more scientific)

9. ### #12 Expert

Nov 30, 2010
18,082
9,647
It's all about your reference point and thinking habits. Pick up a simple schematic and redraw it with negative as the "supply" and go through the process of analysing the circuit under those conditions. After a few of these, you will be able to do it without redrawing the circuit.

Edit: Responding to part of the next post, Yes, some people (like me) do it both ways with no difficulty. That's because the two methods are interchangable for most circuits. Being able to think in either mode, as necessary, is merely a tool that can be mastered rather quickly.

Last edited: Jun 24, 2012
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10. ### WBahn Moderator

Mar 31, 2012
23,157
6,972
Don't spend time memorizing a bunch of different forumulas for this and that. Focus on understanding the fundamentals and being able to apply them with skill and confidence. Then those special case formulas that you end up constructing repeatedly you will naturally start shortcutting and using.

An example is parallel resistors. The general result for the parallel combination of N resistors:

$
\frac{1}{R_{eq}}\;=\;\sum_{i=1}^N}\frac{1}{R_i}
$

You should be comfortable with coming up with this equation on your own given two circuits, one with N parallel connected resistors of arbitrary size and one with a single resistor that is to be sized so as to be equivalent in terms of its voltage vs. current characterstic.

Once you have that down, then you are in a position to use it to solve general problems in which you have parallel resistors (and, with transform methods, general linear impedances). But you will very frequently deal with the situation in which N=2 and the general formula in that case can be rewritten as:

$
R_{eq}\;=\;\frac{R_1R_2}{R_1+R_2}
$

Most people very quickly simply whip out this formula when dealing with two parallel resistors. To the best of my knowledge, it has no one's name attached to it, but whether it does or not it is useful enough that most people commit it to memory through practice.

But what about the equivalent special case for three resistors?

$
R_{eq}\;=\;\frac{R_1R_2R_3}{R_1R_2+R_1R_3+R_2R_3}
$

In fact, if you understand the general case and look for the pattern in these two special cases, you can right down the special case for R=10, or whatever, directly. But how many people ever use them? I'm sure there are a few who, because of the specific nature of what they do day in and day out, it is a oft-useful shortcut. But most people use the special case for N=2 and the general case for everything else.

The same is true for Millman's theorem. It is a direct and readily apparent reorganization of the results obtained for the voltage on a node using Nodal Analysis, which is a more general result. Constraining the general result to one consisting of purely parallel branches in which each branch is a current source or a resistor possibly in series with a voltage source leads directly to MillMan's formulation.

Focus on understanding and being able to apply Nodal and Mesh analysis. You will soon discover that you have developed the ability to write down the necessary equations, by inspection, for commonly occuring special cases and that at least some subset of the topologies covered by Millman's theorem will be in that list.

Almost all scientists and engineers use conventional current exclusively. In nearly all cases it is equivalent and makes no physical difference, but it makes the math a lot cleaner and more consistent. There are exceptions and people that work in those fields commonly think in terms of whichever flow is important for what they do; for some it is electron flow because they are dealing with electron beams, for others it is positive current flow because they are dealing with beams of positively charged ions, others, such as those that work with semiconductors at the lowest levels, learn to think in terms of "majority" and "minority" carriers which could be electrons and "holes" in one region and the reverse, holes and electrons, in another. But unless you end up in one of these specialties, you will be far better served by sticking with conventional current flow.

When all is said and done, the concept of electrical current is a rather abstract one (except for special cases such as those mentioned already). That it is the electrons that are physically moving in the opposite direction of conventional current is a consequence of electrons being assigned as negatively charged. But this was an arbitrary assignment in the first place. Benjamin Franklin, in describing the properties of static electricity, noticed that when you rubbed two suitable pieces of material together they each became statically charged. But he noticed that if you rubbed two other, different materials that also became statically charged and then examined the attraction/repulsion properties of the complete set that charge came in two varieties, which he called positive and negative. He was thinking in terms of some kind of "fluid" and that rubbing materials transfered some of this fluid from one to another and said that the one that had too much fluid was positively charged and the one that had too little fluid was negatively charged. Since he believed that the glass of a Leyden Jar was the holder of the fluid, he assumed that when a glass rod was rubbed with silk that the glass rod got excess fluid was was, hence, positively charged. By the time it was discovered that the mobile charge carrier was the electron and the charge on the glass rod was physically a shortage of electrons, the scientific world had already long since adopted and was using Franklin's definition of what constituted a positive charge.

That all of the observations made by so many were indistinquishable for so many decades underscores the abstractness and arbitrariness of the definition and also that one is no less scientific that the other when working above a certain level.

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11. ### piracyer Thread Starter Member

Jun 9, 2012
32
1
That's great a suggestion.

I think it'd be better for me to able to use the two ways of current flow interchangeably, since different books use different current flows.
Another reason that why I want to get familiar with electron flow is because the online book here is actually using the flow electron.

I kind of get it after spending many hours with the circuit from the worksheet. I figured that when using the flow of electrons, the current from negative to positive is a drop and from positive to negative is a charge as opposed to the other way around in the conventional current. Thus, in the flow of electrons, the negative side of a battery (if there's only one power supply) represents the highest potential and the positive side of the battery represent the lowest potential. Correct me if I am wrong.

Once again, thank you guys for the awesome replies.!!

Last edited: Jun 25, 2012
12. ### WBahn Moderator

Mar 31, 2012
23,157
6,972
As a generalization (and almost certainly too strong a one), texts written from a "technician" perspective tend to me more mixed with many using electron flow and some using conventional flow. I have never seen an "engineering" text that used electron flow (but I'm sure that there have to be some). I think the difference stems from the level of math and theory that each type delves into. In the first type, formulas are often presented with hand-wavy explanations for where they come from and it is easier to visualize the explanations if you picture electrons moving about. In the second type, formulas are the product of pretty rigorous mathematical development from fundamental principles. In fact, it is the nature of the math and not the intended audience that determines which group I would place it in and "technician" and "engineer" are simply convenient qualitative labels. As such, it is very important to use a system that is very mathematically self-consistent, and that means conventional flow.

If you want to use electron flow, then you have to keep a bunch of rules in mind. Yes, for electrons, the negative terminal of the battery is the higher potential (note that this is true for conventional flow if the charge involved is negatively charged) and the electrons move from the higher (negative) potential to the lower (positive) potential. But you are still stuck with the voltages as assigned for conventional flow. So a drop in potential is a gain in voltage. Thus when electron current flows through a resistor, it gains voltage (because losing potential equates to a higher voltage). That's part of what makes working with electron current problematic because you have to apply rules that are not self-consistent. Now, as with anything, once you use it enough it becomes second-nature and the inconsistencies seem to fade away and disappear because you are able to simply deal with them unconsciously.

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